Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. What is the probability that the number will be more than 4 or odd

6 of the 10 integers between 1 and 10 inclusive are more than 4 (that is, 5,6,7,8,9,10).
The probability is the number of favorable outcomes divided by the number of possible outcomes: $P(>4)=\mathrm{\#}$of favorable outcomes $\mathrm{\#}$ of possible outcomes=$\frac{6}{10}$
5 of the 10 integers between 1 and 10 inclusive are odd (that is, 1,3,5,7,9). P(odd)=# of favorable outcomes/# of possible outcomes=$\frac{5}{10}$
3 of the 10 integers between 1 and 10 inclusive are move than 4 and odd (that is, 5,7,9). $P(>4$ and odd$)=\mathrm{\#}$ of favorable outcomes/# of possible outcomes=$\frac{3}{10}$
Use the General addition rule for any two events: $P(AUB)=P(A)+P(B)-P(A\bigcap B)$
P(>4 or odd) $=P(>4)+P(odd)-P(>4\text{and odd})=\frac{6}{10}+\frac{5}{10}-\frac{3}{10}=\frac{6+5-3}{10}=\frac{8}{10}=\frac{4}{5}$